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Croatian Junior Mathematical Olympiad

Geometry Difficulty 4.2 AIME Prove it Croatia

Let $A$ , $B$ , $C$ and $D$ be points on a circle such that $|AB| = |BC| = |CD|$ . The angle bisectors of $\angle ABD$ and $\angle ACD$ intersect at the point $E$ .
If the lines $AE$ and $CD$ are parallel, find $\angle ABC$ .
(Matko Ljulj)

Solution

Due to symmetry, the isosceles triangles $ABC$ and $BCD$ are congruent, and the cyclic quadrilateral $ABCD$ is an isosceles trapezium.

Let us denote by $x$ the measure of angles along the bases in $ABC$ and $BCD$ . Then $\angle ABC = \angle BCD = 180^\circ - 2x$ and $\angle ABD = \angle ACD = 180^\circ - 3x$ , from which we get $\angle EBC = \angle ECB = 90^\circ - \frac{x}{2}$ and $\angle BEC = x$ . Hence, the point $E$ lies on the same circle as $A$ , $B$ , $C$ and $D$ .
We also have $\angle ADC = \angle ADB + \angle BDC = \angle ACB + \angle BDC = x + x = 2x$ and $\angle ECD = \frac{1}{2}\angle ACD = 90^\circ - \frac{3}{2}x$ .
Since $AE \parallel CD$ , the cyclic quadrilateral $ACDE$ is an isosceles trapezium as well, so $\angle ADC = \angle ECD$ holds, and we get $7x = 180^\circ$ . Therefore, $\angle ABC = \frac{5}{7} \cdot 180^\circ$ .

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